Discussion
If \(A \leq 2\sin^{-1}x + \cos^{-1}x \leq B\) , then \(A\) and \(B\) respectively, are
Answer: E
\(A = 0\) and \(B = \pi\).
Use \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\):
\[ 2\sin^{-1}x + \cos^{-1}x = \sin^{-1}x + \frac{\pi}{2} \]
As \(x\) runs over \([-1, 1]\), \(\sin^{-1}x\) runs over \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), so the expression runs from \(-\frac{\pi}{2} + \frac{\pi}{2} = 0\) to \(\frac{\pi}{2} + \frac{\pi}{2} = \pi\). The least and greatest values are 0 and \(\pi\); the other pairs do not match these bounds.
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